Tangents with Parametric Equations

Tangents with Parametric Equations - Tangents with...

Info iconThis preview shows pages 1–4. Sign up to view the full content.

View Full Document Right Arrow Icon
Tangents with Parametric Equations In this section we want to find the tangent lines to the parametric equations given by, To do this let’s first recall how to find the tangent line to at . Here the tangent line is given by, Now, notice that if we could figure out how to get the derivative from the parametric equations we could simply reuse this formula since we will be able to use the parametric equations to find the x and y coordinates of the point. So, just for a second let’s suppose that we were able to eliminate the parameter from the parametric form and write the parametric equations in the form . Now, plug the parametric equations in for x and y . Yes, it seem silly to eliminate the parameter, then immediately put it back in, but it’s what we need to do in order to get our hands on the derivative. Doing this gives, Now, differentiate with respect to t and notice that we’ll need to use the Chain Rule on the right hand side.
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Let’s do another change in notation. We need to be careful with our derivatives here. Derivatives of the lower case function are with respect to t while derivatives of upper case functions are with respect to x . So, to make sure that we keep this straight let’s rewrite things as follows. At this point we should remind ourselves just what we are after. We needed a formula for or that is in terms of the parametric formulas. Notice however that we can get that from the above equation. Notice as well that this will be a function of t and not x . As an aside, notice that we could also get the following formula with a similar derivation if we needed to, Derivative for Parametric E quations
Background image of page 2
Why would we want to do this? Well, recall that in the arc length section of the Applications of Integral section we actually needed this derivative on occasion. So, let’s find a tangent line.
Background image of page 3

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Image of page 4
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

Page1 / 11

Tangents with Parametric Equations - Tangents with...

This preview shows document pages 1 - 4. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online